Question

A mass $m$ is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. The whole system is at rest. At $t=0$, $m$ is released and the system falls freely under gravity. Assume that natural length of the spring is $L_0$, its initial stretched length is $L$ and the acceleration due to gravity is $g$. What is distance between masses as a function of time ?

• A. $L_0+(L-L_0)cos\sqrt{\frac{2k}{m}}t$
• B. $L_0+(L-L_0)cos\sqrt{\frac{k}{m}}t$
• C. $L_0-2+(L+2L_0)cos\sqrt{\frac{2k}{m}}t$
• D. $L_0+(L-L_0)sin\sqrt{\frac{2k}{m}}t$

Answer 1

The correct option is A $L_0+(L-L_0)cos\sqrt{\frac{2k}{m}}t$

In COM frame both the masses execute SHM with
$\omega =\sqrt{\frac{k}{\mu }}$

Here, $\mu =$ reduced mass
$\mu =\frac{m_1\times m_2}{m_1+m_2}=\frac{m^2}{m+m}=\frac{m}{2}$

$\therefore \omega =\sqrt{\frac{2k}{m}}$

Initially, particles are at extreme and distance between them is $L$. Hence, required distance $=L_0+(L-L_0)cos\sqrt{\frac{2k}{m}}t$